Thermodynamics

Enthalpy

100

The sign of delta H for an endothermic reaction.

What is positive?

200

The sign of delta H that is most favorable for a spontaneous reaction.

What is negative?

300

True or False: delta H = H_reactants - H_products

False

400

The technique that can be used to determine the enthalpy of a reaction experimentally.

What is calorimetry?

500

The enthalpy of the dissociation of MgCl₂ in water given the following values.

H_Mg+2 = -462.0 kJ/mol

H_Cl- = -167.4 kJ/mol

H_MgCO3 = -1113 kJ/mol

+483.6 kJ/mol

Entropy

100

Entropy is a measure of the _________ of a system.

Disorder

200

Sign of delta S for the reaction:

2 NI_{3 (s)} --> N_{2 (g)} + 3 I_{2 (g)}

Positive

300

Reaction A is run at 20 ^oC and the same reaction (Reaction B) is run again at 45 ^oC. Which has a greater delta S?

Reaction B

400

Is a positive or negative entropy more favorable?

Positive

500

A reaction is performed at 25 ^oC. If the Gibbs free energy is 250 kJ/mol and the enthalpy is -50 kJ/mol, what is the entropy in kJ/mol x K.

-1.01 kJ/molK

Gibbs Free Energy

100

The sign of delta G for a spontaneous reaction.

Negative

200

If enthalpy is negative and entropy is positive, is the reaction spontaneous or nonspontaneous or depend on temperature?

Spontaneous

300

If enthalpy is negative and entropy is negative, will the reaction be spontaneous, nonspontaneous, or dependent on temperature?

Dependent on temperature

400

Calculate delta G of a reaction at 273 K if the enthalpy is -145 kJ/mol and the entropy is 950 J/mol x K.

-404 kJ/mol OR -404,000 J/mol

500

Can endothermic reactions be spontaneous? If yes, explain why using the signs of delta H, S, and G. If not, give an example of a reaction that is endothermic but spontaneous.

Yes! If the entropy is large enough (positive) to outweigh the unfavorable enthalpy, the reaction will occur spontaneously.

Examples may vary but in class we did dissociation of NH4Cl. Vinegar and Baking soda.

Final Jeopardy

100

Gibbs free energy & spontaneity of the following reaction:

Pb²⁺  +  2I^-  --> PbI₂

Temperature = 298 K

delta S = -58.23 J/molK

H_Pb(2+) = +1.6 kJ/mol

H_I(-) = -55.9 kJ/mol

H_PbI2 = -175.2 kJ/mol

Answer = 55.55 kJ/mol = nonspontaneous

delta S = -0.05823 kJ/mol

delta H = -75.2 - (1.6-(2 x -55.9)) 

           = -75.2 - (1.6 + 111.8)

           = +38.2 kJ/mol

delta G = 38.2 kJ/mol - (298 x -0.05823)

delta G = 38.2 + 17.35 = 55.55 kJ/mol